Question

A man standing South of a lamp post observes his shadow on the horizontal plane to be 24 feel long. On walking Eastwards 300 feet, he finds his shadow as 30 feet. If his height is 6 ft., obtain the height of the lamp above the plane.

Answer 1

PQ = h is the lamp post; AL the man due South of it and of height 6 ft. and AC = 24 ft. is the length of the shadow. After walking 300 ft. Eastwards he comes to B and now the length of shadow is BD =30 ft.
Now let AP = x and BP = y, AC = 24, BD = 30.
From ${\Delta }^{\prime }s$ QPC and LAC
We have $\frac{h}{6}=\frac{24+x}{24}\therefore x=4(h-6)$ ...(1)
Similarly from ${\Delta }^s$ QPD and MBD
$\frac{h}{6}=\frac{y+30}{30}\therefore y=5(h-6)$ ...(2)
From (1) and (2), we get
$h=\frac{24+x}{4}=\frac{y+30}{5}$ $\therefore y=\frac{5}{4}x$.
Again from rt. angled triangle PAB in which
$\angle PAB={90}^o$
$P{B}^2=P{A}^2+A{B}^2$ or $y^2-x^2=(300{)}^2$
Putting for y and x from (1) and (2) in (3), we get
$(h-6{)}^2[25-16]={300}^2={9.100}^2$
$\therefore h-6=100$ or $h=6+100=106ft.$